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Magazine Chapter 15: Q56E (page 875)
Question: Use the simulation (http://openstaxcollege.org/l/16solublesalts) from the earlier Link to Learning to complete the following exercise: Using 0.01 g\(Ca{F_2},\;\)give the \({K_{sp}}\)values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.
Short Answer
In a solution of NaF the solubility product of \({\rm{Ca}}{{\rm{F}}_2}\)is \(3.9 \cdot {10^{ - 11}}\) .
In a solution of KF, the solubility product of \({\rm{Ca}}{{\rm{F}}_2}\)is \(3.9 \cdot {10^{ - 11}}\).
In a solution of \({\rm{NaN}}{{\rm{O}}_3}\)the solubility product of \({\rm{Ca}}{{\rm{F}}_2}\) is \(3.92 \cdot {10^{ - 11}}\).
In a solution of \({\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2}\)the solubility product of \({\rm{Ca}}{{\rm{F}}_2}\)is \(7.6 \cdot {10^{ - 19}}\).
The values of \({K_{{\rm{sp}}}}\)change, as we change soluble salts, because of the common ion effect.
Step by step solution
Find \({K_{sp}}\) for \(Ca{F_2}\)using NaF:
\(\left( {{{\rm{F}}^ - }} \right)\)the initial concentration of \(\left( {{{\rm{F}}^ - }} \right)\)will be 0.2M.
So, the value of x is \(9.75 \cdot {10^{ - 10}}{\rm{mol}}/{\rm{L}}\).
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {C{a^{2 + }}} \right) \cdot {{\left( {{F^ - }} \right)}^2}}\\{ = x \cdot {{(0.2 + 2x)}^2}}\end{array}\)
\(\begin{array}{*{20}{c}}{ = 9.75 \cdot {{10}^{ - 10}} \cdot {{\left( {0.2 + 2 \cdot 9.75 \cdot {{10}^{ - 10}}} \right)}^2}}\\{ = 3.9 \cdot {{10}^{ - 11}}}\end{array}\)
Find \({K_{sp}}\) for \(Ca{F_2}\)using KF:
\(\left( {{{\rm{F}}^ - }} \right)\)the initial concentration of \(\left( {{{\rm{F}}^ - }} \right)\)will be 0.2M.
So, the value of x is \(9.75 \cdot {10^{ - 10}}{\rm{mol}}/{\rm{L}}\).
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {C{a^{2 + }}} \right) \cdot {{\left( {{F^ - }} \right)}^2}}\\{ = x \cdot {{(0.2 + 2x)}^2}}\end{array}\)
\(\begin{array}{*{20}{c}}{ = 9.75 \cdot {{10}^{ - 10}} \cdot {{\left( {0.2 + 2 \cdot 9.75 \cdot {{10}^{ - 10}}} \right)}^2}}\\{ = 3.9 \cdot {{10}^{ - 11}}}\end{array}\)
Find \({K_{sp}}\) for \(Ca{F_2}\)using \(NaN{O_3}\):
The initial concentration of \({\rm{C}}{{\rm{a}}^{2 + }}\)and \({{\rm{F}}^ - }\) is 0 M.
So, the value of x is \(2.14 \cdot {10^4}{\rm{mol}}/{\rm{l}}\)
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {C{a^{2 + }}} \right) \cdot {{\left( {{F^ - }} \right)}^2}}\\{ = x \cdot {{(2x)}^2}}\\{ = 2.14 \cdot {{10}^{ - 4}} \cdot {{\left( {2 \cdot 2.14 \cdot {{10}^{ - 4}}} \right)}^2}}\\{ = 3.92 \cdot {{10}^{ - 11}}}\end{array}\)
Find \({K_{sp}}\) for \(Ca{F_2}\)using \(Ca{\left( {N{O_3}} \right)_2}\):
The initial concentration of \({\rm{C}}{{\rm{a}}^{2 + }}\) is 0.2 M.
So, the value of x is \(9.75 \cdot {10^{ - 10}}{\rm{mol}}/{\rm{L}}\).
\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {C{a^{2 + }}} \right) \cdot {{\left( {{F^ - }} \right)}^2}}\\{ = (0.2 + x) \cdot {{(2x)}^2}}\\{ = \left( {0.2 + 9.75 \cdot {{10}^{ - 10}}} \right) \cdot {{\left( {2 \cdot 9.75 \cdot {{10}^{ - 10}}} \right)}^2}}\\{ = 7.6 \cdot {{10}^{ - 19}}}\end{array}\)
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